Tall towers support power lines h = 56 m above the ground and = 17 m apart that

Question

Tall towers support power lines h = 56 m above the ground and = 17 m apart that run from a hydroelectric plant to a large city, carrying 60 Hz alternating current with amplitude 5 104 A (see figure below). That is, the current in both of the power lines is I = (5 104 A) sin(2 · 60 Hz · t).

(A) Calculate the amplitude (largest magnitude) and direction of the magnetic field produced by the two power lines at the base of the tower, when a current of 5 104 A in the left power line is headed out of the page, and a current of 5 104 A in the right power line is headed into the page.

(b) This magnetic field is not large compared to the Earth's magnetic field, but it varies in time and so might have different biological effects than the Earth's steady field. For a person lying on the ground at the base of the tower, approximately what is the maximum emf produced around the perimeter of the body, which is about 2 m long by half a meter wide?

Explanation / Answer

(a)
The distance of base from power line is

R = sqroot[(l/2)^2 + h^2]
R = sqroot[(17/2)^2 + 56^2]
R = 56.64 m

thita = tan^-1(17/ 56)
thita = 16.88 degree

I(t) = I0*sin(2*pi*f*t)

B(t) = Bout + Bin

Bout(t) = [(u0/4*pi)*(2*Iout(t)/R)]< costhita sinthita 0 >

Bin(t) = [(u0/4*pi)*(2*Iin(t)/R)]< -costhita sinthita 0 >

B(t) = 2*[(u0/4*pi)*(2*I(t)/R)]*sinthita y^

B(t) = 2*[(u0/4*pi)*(2*(5*10^4)/56.64)]*sin(16.88) y^

B(t) = 1.025*10^4 T

(b)
The induced emf is

B(t) = 2*[(u0/4*pi)*(2*I(t)/R)]*sinthita y^

I(t) = I0*sin(2*pi*t*f)

emf = -4*[(u0/4*pi)*(Abody*sinthita /R)*(dI(t)/dt)

emf = -4*[(u0/4*pi)*(Abody*sinthita /R)*(I0*2*pi*f*cos(2*pi*f*t))

emfmax. = [(Abody*sinthita /R)]*(u0*I0*2*f)

= [2*0.5*sin(16.88)/59.5]*(u0*5*10^4*2*60)

= 36.79 mV

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.